Third Normal Form Decompositions
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Given a relation R and a minimal set F of fds is not in 3NF,
we compute the following decomposition:

     - Create a new relation for each functional dependency
     - If no relation has all the attributes in one of the keys, then
     add one more relation with attributes of one of the keys
     - If one relation has a subset of the attributes of another relation
     then remove the smaller relation.

3NF decomposition is:
     - it is always lossless
     - it is dependency preserving.
     - all resulting relations are in 3NF

Ex:

R(A,B,C,D,E,F)
AB->C
CD->EF
F->A

Keys: ABD, BDF

3NF decomposition:

R1(A,B,C)      F1 = {AB->C}   Key: AB
R2(C,D,E,F)    F2 = {CD->EF}  Key: CD
R3(A,F)        F3 = {F->A}    Key: F
R4(A,B,D)      F4 = {}        Key: ABD


S(A,B,C,D)
AB->C
A->D
C->A
Key:AB

S1(A,B,C)  F1 = {AB->C, C->A}   Keys: AB, BC  (in 3NF, not in BCNF)
S2(A,D)    F2 = {A->D}          Keys: A   (in 3NF, BCNF)